Chapter 14Homework 14A Page 448: 71-76, 78, 79, 88-90

71. State the laws of Boyle, Charles, and Gay- Lussac as equations. Boyles Law: P1V1 = P2V2; Charless Law: V1/T1 = V2/T2; Gay-Lussacs Law: P1/T1 = P2/T2 72. What will happen to the pressure of a contained gas if its temperature is lowered? Pressure decreases.

73. Why is it important not to puncture an aerosol can? Aerosol cans contain gases under pressure that can expand explosively if allowed to escape too rapidly. 74. Explain why an unopened bag of potato chips left in a hot car appears to become larger. The air trapped inside the bag expands as it gets hotter because volume is directly proportional to temperature.

75. If two variables have an inverse relationship, what happens to the value of one as the value of the other increases? In an inverse relationship, one variable always decreases as the other increases. 76. If two variables have a direct relationship, what happens to the value of one as the value of the other is increased? In a direct relationship, one variable always increases as the other increases.

78. Label the following examples as being representative of a direct or inverse relationship. a. pressure versus volume of a gas inverse b. volume versus temperature of a gas direct c. pressure versus temperature of a gas direct

79. What four variables are used to describe gases? volume, pressure, temperature, and number of moles 88. Use Boyles, Charless, or Gay-Lussacs law to calculate the missing value in each of the following. a. V1 = 2.0 L, P1 = 0.82 atm, V2 = 1.0 L, P2 = ?

P1V1 Since P1V1 ! P2V2 , then P2 ! V2 P1V1 (0.82atm)(2.0 L) P2 ! ! ! 1.6 atm V2 1.0L

88. b. V1 = 250 mL, T1 = ?, V2 = 400 mL, T2 = 298 K

V1 V2 since ! , then ... T2V1 ! T1V2 , and... T1 T2

T2V1 (298K)(250.mL) T1 ! ! ! 186K V2 400mL(or 200 K, if you consider 400 mL = 1 SF!)

88. c. V1 = 0.55 L, P1 = 740 mm Hg, V2 = 0.80 L, P2 ?

P1V1 (740mmHg)(0.55 L) P2 ! ! ! 510mmHg V2 0.80Ld. T1 = 25C, P1 = ?, T2 = 37C, P2 = 1.0 atm T1= 25C + 273 = 298K, and T2 = 37C +273 = 310K

P1 P2 T 1P 2 Since ! , then P1 ! , and... T 1 T2 T2 T1P2 (298K)(1.0atm) P1 ! ! ! 0.96atm T2 310K

89. What is the pressure of a fixed volume of a gas at 30.0C if it has a pressure of 1.11 atm at 15.0C? T1 = 15.0C +273 = 288K, T2 = 30.0C +273 = 303 K

P1 P2 T 2P 1 Since ! , then P2 ! , and... T1 T2 T1 T2P1 (303K)(1.11atm) ! ! 1.17atm P2 ! 288K T1

90. A fixed amount of oxygen gas is held in a 1.00-L tank at a pressure of 3.50 atm. The tank is connected to an empty 2.00-L tank by a tube with a valve. After this valve has been opened and the oxygen is allowed to flow freely between the two tanks at a constant temperature, what is the final pressure in the system? The key to this problem is that the total volume of the gas has been increased from 1.00-L to 3.00-L (1.00 + 2.00 L). Of course, the pressure will decrease!

90. A fixed amount of oxygen gas is held in a 1.00-L tank at a pressure of 3.50 atm. The tank is connected to an empty 2.00-L tank by a tube with a valve. After this valve has been opened and the oxygen is allowed to flow freely between the two tanks at a constant temperature, what is the final pressure in the system?

P1V1 Since P1V1 ! P2V2 , then P2 ! V2 P1V1 (3.50atm)(1.00 L) P2 ! ! ! 1.17 atm V2 3.00L

Chapter 14Homework 14B Pages 448, 449: 80-83, 92-96

80. List the standard conditions for gas measurements. T = 0.00C = (273K) and P = 1.00 atm (760 mmHg) 81. Write the equation for the combined gas law. Identify the units most commonly used with each variable.

P1V1 P2V2 ! T1 T2

Units commonly used for the variables are atm, mmHg (torr), and kPa for pressure, Kelvin for temperature, and L for volume.

82. State Avogadros principle. At a fixed temperature and pressure, equal volumes of any gas contain equal numbers of particles. 83. What volume is occupied by one mole of a gas at STP? What volume do two moles occupy at STP? 22.4 L; 44.8 L

92. A sample of nitrogen gas is stored in a 500.0-mL flask at 108 kPa and 10.0C. The gas is transferred to a 750.0-mL flask at 21.0C. What is the pressure of nitrogen in the second flask? T1 = 10.0C +273 = 283K, T2 = 21.0C +273 = 294K V1 = 500.0-mL, V2 = 750.0-mL, P1 = 108 kPa, P2 =?

P1V1 P2V2 T2P1V1 If ! , then ! P2 T1 T2 V2T1 T2P1V1 (294K)(108kPa)(500.0mL) ! P2 ! V2T1 (750.0mL)(283K)P2 = 74.8 kPa

93. The air in a dry, sealed 2-L soda bottle has a pressure of 0.998 atm at sea level at a temperature of 34.0C. What will be its pressure if it is brought to a higher altitude where the temperature is only 23.0C? T1 = 34.0C + 273 = 307K, T2 = 23.0C + 273 = 296K Volume is held constant. V1 = V2 (V drops out) P1 = 0.998atm, P2 = ??

P1 P2 T 2P 1 Since ! , then P2 ! , and... T 1 T2 T1 T2P1 (296K)(0.998atm) P2 ! ! ! 0.962atm T1 307K

94. A weather balloon is filled with helium that occupies a volume of 5.00 x 104 L at 0.995 atm and 32.0C. After it is released, it rises to a location where the pressure is 0.720 atm and the temperature is -12.0C. What is the volume of the balloon at that new location? T1 = 32.0C + 273 = 305K, T2 =-12.0C + 273 = 261K P1 = 0.995atm, P2 = 0.720atm, V1 = 5.00E4L, V2 = ?

P2V2 P1V1 T2P1V1 , then If ! ! V2 T2 T1 P2T1 T2P1V1 (261K)(0.995atm)(5.00E4L) ! V2 ! P2T1 (0.720atm)(305K) V2 = 5.91 x 104 L

95. Propane, C3H8, is a gas commonly used as a home fuel for cooking and heating. a. Calculate the volume that 0.540 mol of propane occupies at STP.

22.4L 0.540mol x ! 12.1Liters (@ STP) 1 molb. Think about the size of this volume compared to the amount of propane that it contains. Why do you think propane is usually liquefied before it is transported? Propane occupies a much smaller volume when liquefied. (0.540 mol propane = only 23.8 grams!)

96. Carbon monoxide, CO, is a product of incomplete combustion of fuels. Find the volume that 42 g of carbon monoxide gas occupies at STP.

1mol CO 22.4 L 42g CO x x ! 28.0g CO 1 mol

34 Liters (@ STP)(2 sig figs!)

Chapter 14Homework 14C Page 437: 41, 43-45 Page 449: 97-100

41. If the pressure exerted by a gas at 25C in a volume of 0.044L is 3.81atm, how many moles of gas are present? n = ?, T = 25C + 273 = 298K, V = 0.044L, P = 3.81 atm and since Pressure is in atm, R = 0.0821 Latm/Kmol

PV If PV ! nRT, then n ! RT PV (3.81atm)(0.044L) ! ! n! L yatm RT (0.0821 K ymol )(298K)= 6.9 x 10-3 mol (2 sig figs)

43. Calculate the volume that a 0.323-mol sample of a gas will occupy at 265 K and a pressure of 0.900 atm. n = 0.323mol, T = 265K, V = ? L, P = 0.900 atm and since Pressure is in atm, R = 0.0821 Latm/Kmol

nRT If PV ! nRT, then V ! P Lyatm nRT (0.323mol)(0.0821 K ymol )(265K) ! ! V! P (0.900atm)V = 7.81 Liters

44. What is the pressure in atmospheres of a 0.108-mol sample of helium gas at a temperature of 20.0C if its volume is 0.505 L? n = 0.108mol, T = 20C + 273 = 293K, V = 0.505L, P = ? atm (P in atm, so) R = 0.0821 Latm/Kmol

nRT If PV ! nRT, then P ! V Lyatm nRT (0.108mol)(0.0821 K ymol )(293K) ! ! P! V (0.505L)= 5.14 atm

45. Determine the Kelvin temperature required for 0.0470 mol of gas to fill a balloon to 1.20 L under 0.988 atm pressure. n = 0.0470mol, T = ?, V = 1.20L, P = 0.988atm (P in atm, so) R = 0.0821 Latm/Kmol

PV If PV ! nRT, then T ! nR PV (0.988atm)(1.20L) ! ! T! L yatm nR (0.0470)(0.0821 K ymol )T = 307K

97. The lowest pressure achieved in a laboratory is about 1.0 x 10-15 mm Hg. How many molecules of gas are present in a 1.00-L sample at that pressure and a temperature of 22.0C? n = ?, T = 22C + 273 = 295K, V = 1.00L, P = 1.0E-15mmHg, so R = 62.4 LmmHg/Kmol

PV If PV ! nRT, then n ! RT PV (1.0E - 15 mmHg)(1.00L) n! ! ! Ly mmHg RT (62.4 K ymol )(295K) 6.02E23 molecules 5.4E 20mol x ! 3.3E4 1 mol

98. Determine the density of chlorine gas at 22.0C and 1.00 atm pressure. Assume 1 mol of Cl2(g). (Molar mass is grams per mol and per means for each 1) Or use Molar mass of Cl2 = 71.0 g/mol (D) = M P T = 22.0C + 273 = 295K R T solve for V, then find Density

)(295K) nRT (1.00mol)(0.0821 ! ! V! P (1.00atm) mass 71.0 g ! ! 2.93g/L V= 24.2 L D ! vol 24.2L

Lyatm K y mol

99. Geraniol is a compound found in rose oil that is used in perfumes. What is the molar mass of geraniol if its vapor has a density of 0.480 g/L at a temperature of 260.0C and a pressure of 0.140 atm? Assume 1 L of geraniol. T = 260.0C + 273 = 533K Find moles, then use g/mol to find molar mass

PV (.140 atm)(1 L) ! ! n! Lyatm RT (533K)(0.0821 K ymol )(n = 3.19E-3 mol for 1 L) M !

0.480g ! 150.g/mol 3.19E - 3 mol

100. A 2.00-L flask is filled with propane gas (C3H8) at 1.00 atm and 15.0C. What is the mass of the propane in the flask? T = -15C + 273 = 258K, find n and convert to g.

PV (1.00atm)(2.00L) n! ! ! Lyatm RT (0.0821 K ymol )(258K)n = 0.0944 molC3H8 = 3(12.0) + 8(1.0) = 44.0 g/mol

44.0g 0.0944 mol x ! 4.15 g 1mol